Cyclic groups

**dimuk** · March 20th 2009, 12:43 AM

Let G=Sym(n) .Let C be a cyclic group. Suppose that $\theta \in Hom(C, G).$ then how to show that the generators for $\theta (C)$ in G all have the same cycle type and so are conjugate in G.

**ThePerfectHacker** · March 20th 2009, 08:36 AM

Originally Posted by dimuk

Let G=Sym(n) .Let C be a cyclic group. Suppose that $\theta \in Hom(C, G).$ then how to show that the generators for $\theta (C)$ in G all have the same cycle type and so are conjugate in G.

I am assuming that the cyclic group is finite.

If $a$ generates $\theta (C)$ then $a^j$ generates $\theta (C)$ where $j$ is relatively prime with the order of $a$ because a homomorphic image of a cyclic group is cyclic. Let $a=\sigma_1\sigma_2 ... \sigma_k$ for disjoint cycles $\sigma_i \in S_n$ . The order of $a$ is the least common multiple of the lengths of all these cycles. Therefore, $|\sigma_i|$ divides $|a|$ for all $i$ . Now if $j$ is an integer so that $(j,|a|) = 1$ then certainly $(j,|\sigma_i|)=1$ . Therefore, if $\sigma_i$ is an $i$ -cycle then $\sigma_i^j$ still stays as an $i$ -cycle. Thus, we have $a^j = \sigma_1^j ... \sigma_k^j$ where $\sigma_i$ is same cycle length as $\sigma_i^j$ which means $a$ and $a^j$ have the same cycle type and consequently they are conjugate.

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